Sleeping Beauty and Tuesdays

I am trying to make a point that, mathematically, the Sleeping Beauty problem is resolved, and the Wikipedia article about it should stop assuming that there is an ongoing debate.

The Sleeping Beauty problem. Sleeping Beauty participates in the following experiment. On Sunday, she is put to sleep, and a fair coin is flipped. Regardless of the result of the coin flip, she is awakened on Monday and asked whether she thinks the coin was heads or tails. Regardless of her answer, if the coin was heads, the experiment ends. However, if the coin was tails, she is put back to sleep with her memory erased and awakened again on Tuesday and asked the same question. In this case, the experiment stops on Tuesday. She knows the protocol. She is awakened one morning. From her point of view, what is the probability that the coin was heads?

Here is the solution. If it is Monday, then the probability that the coin is heads is one half. So the probability of Monday/heads is the same as Monday/tails. If the coin is tails, Sleeping Beauty can’t distinguish between Monday and Tuesday. So the probability of Monday/tails is the same as Tuesday/tails. Thus, the three cases Monday/heads, Monday/tails, and Tuesday/tails are equally probable. It follows that when she is awakened, the probability of heads is one third.

However, there are still people — called halfers — who think that the probability of heads is one half.

But suppose we ask a different question.

Different question. She is awakened one morning. From her point of view, what is the probability that the day is Tuesday?

As I explained before, when she is awakened, the probability of it being Tuesday is one third. Let us calculate what the halfers think. Suppose they think that the probability of the day being Tuesday is x, then the probability of Monday is 1 − x. Let’s calculate the probability of the coin being heads from here. The probability of heads, if today is Tuesday, is zero. The probability of heads, if today is Monday, is 1/2. Therefore, the probability of heads equals 0 · x + (1 − x)/2 = (1 − x)/2. In halfers’ view, the resulting calculation equals 1/2. In other words, (1 − x)/2 = 1/2. It follows that x is zero. Doesn’t make much sense that the probability of the day being Tuesday is zero, does it?

Halfers are wrong. Wikipedia should update the article.



  1. Jeff Jo:

    Tanya, while I agree 100% with you, there is a better solution. The issue is that it is the setup for the solution is debated, now how that solution unfolds. So what is needed is a better setup.

    Go back and read Elga. The original problem did not mention Monday, Tuesday, or an ordering of the mandatory awakening (i.e., Monday) and the optional one. Elga introduced those elements as part of his solution, which is basically the one you describe. But the problem, with irrelevant details redacted, actually is: “Some researchers are going to put you to sleep. During the [experiment], they will briefly wake you up either once or twice, depending on the toss of a fair coin (Heads: once; Tails: twice). After each waking, they will put you to back to sleep with a drug that makes you forget that waking. When you are … awakened, to what degree ought you believe that the outcome of the coin toss is Heads?”

    Here’s a better way to set it up, consistent with both this Elga’s stup, as well as his solution. But it has a simple solution:

    1) After SB is asleep, flip two coins instead of one. Call them C1 and C2.
    2) If either is showing Tails, wake SB and interview her. After she answers th interview question, put her back to sleep with amnesia.
    3) Turn C2 over.
    4) Repeat step 2.

    In the interview, as SB for her degree of belief that C1 is currently showing Heads.

    Solution: Regardless of whether this is step 2 or step 4, at the beginning of the step (while SB was still asleep) there were four equally-likely combinations for C1 and C2: HH, HT, TH, and TT. Given the evidence that she was wakened, SB can eliminate HH. In only one of the three remaining cases is C1 showing Heads.

    The problem with Elga’s setup, is that it does not establish a consistent prior for the evidence that SB is awake. Elga argued, as you do, that is is the situation on Monday Night and applies to any awakening, regardless of the differing circumstances known to apply to the different days. Halfers essentially argue, whether they realize it or not, that by not knowing which day it is, being awake ceases to be evidence. Unless you can establish why it is, you can’t convince them that your solution, which assumes it is, is better.

    But my solution establishes a clear prior, and establishes that being awake is evidence.

  2. Jeff Jo:

    Oops – I hit “submit” before proofing. The obvious typos you can correct. But “that is is the situation on Monday Night and applies to any awakening” should be “that IT is the situation on SUNDAY Night and applies to any awakening.”

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